### 2. Graph Coloring

```Graph Coloring
prepared and Instructed by
Shmuel Wimer
Eng. Faculty, Bar-Ilan University
March 2014
Graph Coloring
1
Vertex Coloring
A k-coloring of a graph G is a labeling f : V(G) → { 1,…,k }.
A coloring is proper if no two vertices x and y, connected
with an edge have same color, xy ϵ E(G) => f(x) ≠ f(y).
G is k-colorable if it has proper k-coloring.
The chromatic number χ(G) is the smallest k such that G
has proper k-coloring. G is called k-chromatic.
If χ(G) = k, but χ(H) < k for every proper subgraph H, then
G is k-critical.
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Graph Coloring
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The vertices having same color in a proper k-coloring
must be independent. Therefore, χ(G) is the minimum
number of independent sets covering G.
Hence, G is k-colorable iff G is k-partite.
Examples. Every bipartite graph is 2-colorable.
Every even cycle graph is 2-colorable (it is bipartite).
Every odd cycle graph is 3-colorable and 3-critical.
2-colorability can be tested with BFS. (how?)
We compute the distance from a vertex u. A connected
graph is bipartite iff G[X] and G[Y] are independent
sets, where X and Y are vertices of even and odd
distance from u, respectively.
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Graph Coloring
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The largest clique size ω(G) satisfies χ(G) ≥ ω(G).
The largest independent set size α(G) satisfies
χ(G) ≥ n(G) / α(G).
Is it possible that
χ(G) > ω(G)?
s
Yes, χ(G) may exceed
ω(G)!
Proper coloring of Ks
requites s colors.
G
s s
C5
s s
Ks+2
Ks
Ks with any two adjacent of C5 vertices turn into Ks+2.
χ(G) = s+3 whereas ω(G) = s+2! Hence χ(G) > ω(G).
Could it be constructed with C3 rather than C5?
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Graph Coloring
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Upper Bounds of Chromatic Number
Easy bounds are χ(G) ≤ n(G), χ(G) ≥ ω(G), and
χ(G) ≥ n(G) / α(G), all hold with equality for cliques.
Better than χ(G) ≤ n(G) upper bounds can be obtained
by coloring algorithms.
A greedy algorithm w.r.t V(G) order v1,…,vn , assigns to vi
the smallest color index not incident so far to vi.
Proposition. There is χ(G) ≤ ∆(G) + 1. (∆(G) is the largest
vertex degree.)
Proof. By construction. A vertex has no more than ∆(G)
neighbors. Upon vi coloring there must be at least one
of 1,…, ∆(G)+1 colors unused. ■
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Graph Coloring
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Different orderings may yield smaller upper bounds.
Finding the best ordering is hard. Is there an ordering
yielding χ(G)? It can be shown that such one exists.
Example. Register allocation and interval graphs.
Consider the registers used by a compiler, each has start
and end time. What is the smallest number of physical
registers that can be used?
Assign the symbols a, b, c, … to the registers in the code,
and draw their usage time intervals.
a
b
f
c
d
e
g
Proposition. If G is an interval graph then χ(G) = ω(G).
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Graph Coloring
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Proof. By left-to-right traversal of the time intervals, pre
sorted by their starting time. Initializing k=0. Increasing
to k+1 at starting point and decreasing to k-1 at ending
point. ■
The bound χ(G) ≤ ∆(G)+1
may still be very poor.
For (n+1)-vertex star graph
∆(G) = n, whereas χ(G) = 2.
For (n+1)-vertex wheel
graph ∆(G) = n, whereas
χ(G) ≤ 4.
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Graph Coloring
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The bound χ(G) ≤ ∆(G)+1 can be further improved by
considering the vertices with high degree first.
Proposition. (Welsh-Powell 1967) If the vertices are
ordered in non increasing degree, d1 ≥ d2 ≥ … dn, then
χ(G) ≤ 1 + maxi min { di , i-1 }.
Proof. When vertex i is colored, at most min { di , i-1 } of
its neighbors have already been colored.
Its color is therefore 1 + min{ di , i-1 }. Maximization over
i yields the upper bound. ■
The minimum degree δ(G) in G can also be used to
deduce upper bounds.
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Graph Coloring
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Lemma. If H is k-critical graph, then δ(H) ≥ k-1.
Proof. Assume in contrary that δ(H) < k-1. Let x ϵ H be
any vertex.
Since H is k-critical, H-x is (k-1)-colorable. If dH(x) < k-1,
then not all the k-1 colors used in H-x are used for N(x).
We could therefore drop one of the missing colors for x.
That holds for any x. By color renaming we can obtain a
proper k-1 coloring of H, hence a contradiction. ■
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Corollary. (Szekeres-Wilf 1968)   G   1  m ax H  G   H  .
Proof. Let k    G  and H’ be a k-critical subgraph of G.
By the above lemma   H    k  1    G   1 .
There is also
bound. ■
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  H    m ax H  G   H
Graph Coloring
 , yielding the desired
10
Coloring of Directed Graphs
Theorem. Let a graph G be directed with longest path
l(G), then χ(G) ≤ 1 + l(G). Furthermore, there are
orientations of G’s edges such that equality holds.
2
G
4
3
1
5
G’
6
1
Proof. Let G’ be a maximal acyclic sub digraph of G (not
necessarily a tree).
G’ must have vertices with outgoing arcs only. Define f(v)
to be a coloring function assigning color 1+l(v).
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Graph Coloring
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f is strictly increasing
along a path in G’
using colors 1+l(G’)
on V(G) = V(G’).
For each edge uv ϵ E(G) there exists a path in G’ between
u and v, since either there was uv ϵ E(G’) or the edge is
closing a cycle of G.
That implies f(u) ≠ f(v) since f increases along paths of G’.
Consequently, f is a proper coloring and χ(G) ≤ 1 + l(G).
To show orientations of G’s edges satisfying χ(G) = 1 +
l(G), an orientation satisfying χ(G) ≥ 1 + l(G) is shown.
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Let f be an optimal coloring satisfying f(G) = χ(G). We
derive digraph G* as follows.
Each edge uv ϵ E(G*) is oriented u → v iff f(u) < f(v).
Since f is a proper coloring, this defines an orientation.
Since the color labels
along paths in G* strictly
increase, and there are
only χ(G) labels, there is
l(G*) ≤ χ(G*) – 1, hence
χ(G) = 1 + l(G). ■
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1
2
G
G*
4
Graph Coloring
3
13
Brook’s Theorem
The bound χ(G) ≤ ∆(G) + 1 holds for any graph. Brook
showed that cliques and odd cycles are essentially the
only graphs where χ(G) = ∆(G) + 1 holds.
Theorem. (Brook 1941) If G is connected and other than
a clique or an odd cycle, then χ(G) ≤ ∆(G).
Proof. Let G have n nodes and be connected, neither a
clique, nor an odd cycle. Let k = ∆(G) and assume k ≥ 3,
as otherwise for k = 1 it is a clique, and bipartite or an
odd cycle for k = 2.
Consider first the case where G is not k-regular. Choose
a vertex vn for which d(vn) < k and grow a spanning tree
rooted at vn (by any search).
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Graph Coloring
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Index the vertices in decreasing order as they are being
reached by the search, yielding the order v1, v2, … ,vn.
Each vertex other than vn has a higher-indexed neighbor
along its path to root, hence it has at most k-1 lowerindexed neighbors.
Using the greedy coloring with the vertex decreasing
order obtains proper k-coloring.
In the remaining cases G is k-regular.
1st case: G is 1-connected. Let x be a cut-vertex. Let G’
be a component of G-x together with x. The degree of x
in G’ is less than k and a proper k-coloring is possible.
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Graph Coloring
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G’
x
G
This can be repeated for every components of G-x,
yielding k-proper coloring for each (x included).
By permuting colors within the subgraphs, we can make
the colorings agree on x, yielding k-proper coloring of G.
2nd case: G is not 2-connected. Find a vertex vn with
neighbors v1 and v2 (non adjacent ) whose deletion
leaves a connected subgraph.
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v1
v2
vn
G - {v1 , v2 } is connected and a spanning tree rooted at
vn can be constructed. The labels 3 , … , n are assigned
to the vertices in decreasing order as they are reached.
Starting coloring from v1 and v2, they use same color.
Then, each vertex other than vn has at most k-1 lowerindexed neighbors and k colors can be used for those.
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Graph Coloring
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vn has k neighbors, of which v1 and v2 already used the
same color. The rest neighbors used at most other k-2
colors, and vn can therefore be properly colored.
All in all, k proper coloring of G has been obtained.
3rd case: G is 2-connected. Choose a vertex x such that
vertex connectivity κ(G-x) = 1.
x = vn
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v1
v2
x = vn
x has a neighbor in every block of G-x obtained by
deleting the 2nd vertex in a cut-set, else G is 1-connected
rather than 2-connected.
There is no edge connecting v1 and v2 since they are in
different blocks.
G - {x , v1 , v2 } is connected since blocks have no cutvertices and v1 , v2 are no such.
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k ≥ 3 implies G - {v1 , v2 } is also connected.
All in all this is the same situation as the case of G not 2connected. ■
Brook’s Theorem implies that the cliques and the odd
cycles are the only (k-1)-regular k-critical graphs.
(homework)
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Graph Coloring
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Chromatic Polynomials
We shall associate with any graph a function telling
whether or not it is 4-colorable.
This study was motivated by the hope to prove the FourColor Theorem, which by that time was a conjecture.
Let PG(k) denote the number of proper colorings of a
graph G with k colors. PG(k) is called the chromatic
function of G.
Example. PG(k) = k (k - 1)2. The first
vertex can be colored in k ways, while
each of the other two in k-1 ways.
For a tree T of n vertices there is PT(k) = k (k - 1)n-1.
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For G = K3 there is PG(k) = k (k - 1) (k - 2) and for G = Kn
there is PG(k) = k (k - 1) (k - 2) · · · (k – n + 1).
If k < χ(G) then PG(k) = 0. For k ≥ χ(G) there is PG(k) > 0.
The Four-Color Theorem for planar graph G is equivalent
to the statement that PG(4) > 0.
It is difficult to compute PG(k) by inspection, but it can
be systematically obtained as a sum of chromatic
functions of complete graphs.
Theorem. Let u,v ϵ V(G) be not incident, G1 be obtained
by adding the edge uv to G, and G2 be obtained by
identifying u and v in G. Then PG(k) = PG1(k) + PG2(k).
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Graph Coloring
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u
v
v
G1
G
k(k-1)(k-2)2
u
=
uv
G2
k(k-1)(k-2)(k-3) + k(k-1)(k-2
Proof. In a proper coloring of G, u and v may have either
the same color or different colors.
The number of proper colorings where u and v have
different colors does not change if an edge uv would
exist, yielding PG1(k).
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Graph Coloring
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Similarly, the number of proper colorings where u and v
have same color does not change if an u and v are
merged, yielding PG2(k). ■
Corollary. The chromatic function is a polynomial.
Proof. The procedure of the theorem results in two
graphs. In G1 the number of edges is increased. In G2 the
number of vertices is decreased.
The process is finite. It ends with producing complete
graphs, whose chromatic functions are polynomial.
The chromatic function is therefore a finite sum of
polynomials, which must be polynomial too. ■
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For n-vertex graph G the degree of PG(k) is n,
the coefficient of kn is 1 and that of kn-1 is |E(G)|,
the sign of the coefficients is alternating, and
the free coefficient is zero.
Example.
PG(k) = k5 - 7k4 + ak3 - bk2 + ck
w
v
w
=
+
w
v
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v
Graph Coloring
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v
v
+
=
w
=
+
+
+
w
+2
+
PG(k) = k(k - 1)(k - 2)(k - 3)(k - 4) + 3k(k - 1)(k - 2)(k - 3)
+ 2k(k - 1)(k - 2) = k5 - 7k4 + 19k3 - 23k2 + 10k
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Graph Coloring
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Example: Scheduling feasibility.
A schedule for certain lectures is required, for which
some time slots are given (e.g., campus is open). There
is no limit of available rooms.
It is known that some lectures cannot take place in
parallel (e.g. students are registered to both).
Is scheduling feasible? How many schedules there are?
Solution. Define a graph where lectures are vertices and
edges correspond to lectures that cannot be scheduled
at the same time slot.
The chromatic polynomial, where k is the number of
time slots answer the questions. ■
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Graph Coloring
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Edge Coloring
A k-edge-coloring of a graph G is
a labeling f : E(G) → { 1, … ,k }.
Edge coloring partitions E(G) into
k sets (some possibly empty) { E1,
E2, … , Ek }.
An edge coloring is proper if adjacent edges have
different colors. All coloring henceforth are meant to be
proper.
Edge coloring thus partitions E(G) into k sets
{ M1, M2, … , Mk } of matchings. (Only loopless graphs
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G is k-edge-colorable if it has k-edge-coloring.
Clearly, G is m-edge-colorable, where m = |E(G)|.
The edge chromatic number is the smallest k such that G
has k-edge-coloring. G is called k-edge-chromatic.
Not 3-edge-colorable, hence χ’(G) = 4.
Clearly, χ’(G) ≥ Δ(G).
Example. (Timetabling) m teachers x1, x2, … , xm and n
classes y1, y2, … , yn are given. Teacher xi is required to
teach class yj a lessen of period pij. Schedule a complete
timetable having minimum duration.
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Solution. The scheduling is represented by a bipartite
graph H[X,Y], X = { x1, x2, … , xm } , Y = { y1, y2, … , yn }
vertices xi and yj are connected with pij parallel edges.
The minimum number of colors required for H edgecoloring ensures minimum duration.
• No schedule overlap for a teacher.
• No two lesson schedules overlap in a class. ■
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Graph Coloring
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Edge Coloring of Bipartite Graphs
Let the subgraph H span G (V(H) = V(G)), and
C := { M1, M2, … , Mk } be a k-edge-coloring of H.
A color is available for an e ϵ E(G) \ E(H) if it is available
in its two end vertices.
If e is uncolored, any of its available colors can be
assigned to extend C to a k-edge-coloring of H + e.
For i ≠ j , each component of Hij := H[Mi U Mj] is either an
even cycle or a path (called ij-path). (why?)
Theorem. If G is bipartite then χ’(G) = Δ(G).
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Proof. By induction on m = |E(G)|. Let e = uv ϵ E(G).
Assume that H = G \ e has a Δ-edge-coloring { M1, M2, … ,
MΔ }.
If a color is available for e we are done. Otherwise, each
of the Δ colors is represented either at u or at v.
Since the degrees of u and v in G \ e are Δ-1 at most,
there are colors i ≠ j, where i is available at u and exists in
v, and j is available at v and exists in u.
Consider the subgraph Hij := H[Mi U Mj]. Because u has a
degree one in Hij, the component containing u is an ijpath P.
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P cannot terminate at v. If it did, it would started from u
with color j and end at v with color i, hence comprising
even number of edges.
u
P
P + e would then be an odd
cycle in G, impossible for a
bipartite graph.
v
Interchanging the colors of
u
P
e
P, a new Δ-edge-colorable
H is obtained, where color j
is available at u and v.
v
Assigning color j to e obtains a Δ-edge-coloring of G. ■
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Graph Coloring
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Clearly, there is χ’(G) ≥ Δ(G), and for bipartite graphs
there is χ’(G) = Δ(G). What can be said about an upper
bound? Surprisingly, it is very tight.
Theorem. (Vizing 1964, Gupta 1966). Let G be a simple
(no parallel edges, loopless) graph. Then χ’(G) ≤ Δ(G) + 1.
Proof. Let G’, a proper subgraph of G, be edge-colored
with Δ(G) + 1 colors, but uv could not be colored. We
present a recoloring procedure to include uv.
Since more than Δ(G) colors are used,
every vertex has a color not presented.
Let a0 be missing from u and a1 from v.
(a0 must be presented at v and a1 at u.)
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Graph Coloring
u
v
34
Let v1 be neighbor of u along
the edge colored a1. At v1 some
color a2 must be missing.
Suppose a2 does not appear on
u. We could recolor uv1 with a2,
free a1 from u, and then color uv
with a1.
u
v1
v
u
v
So we suppose that a2 appears on u.
v1
The process continues for i ≥ 2.
Finding a new color ai that appears at u, let vi be the
neighbor of u along the edge colored ai.
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At vi some color ai+1 must be missing.
If ai+1 is missing at u, we downshift color aj from uvj to
uvj-1 for 1 ≤ j ≤ i (uv0 = uv).
vi
ai+1
v2
u
v1
v
vi
vi
ai+1
ai+1
v2
u
v1
v
v2
u
v1
v
We are finished, unless ai+1 appears at u, in which case
the process continues.
There are only Δ(G) + 1 colors, hence the iterative
selection of ai+1 eventually repeats a color.
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Let a1 , … , al be the shortest non repetitive color list such
that al+1 is missing at vl and repeats one of a1 , … , al-1 .
Let al+1 = ak for some 1 ≤ k ≤ l-1.
This color is missing from vk-1 and
appears on uvk.
vk-1
If vl lacks a0 , we downshift colors
from vl and use a0 on uvl to
complete the augmentation.
v2
vk
P
vl
u
v1
v
Hence we assume that a0 appears at vl and ak does not.
Let P be the longest alternating path of edges colored a0
and ak that begins at vl (with a0).
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P is unique. (why?)
Depending on the opposite end of P, recoloring can take
place to complete the augmentation. Three possibilities
exist: P ends at vk, P ends at vk-1, and P ends elsewhere.
If P ends at vk, it is
with a0 since uvk is
colored with ak.
Downshifting from
vk, switching colors
of P, and coloring uvk
with a0 , completes
the augmentation.
March 2014
a0
ak
vk-1
a0
vk
ak
a0
vl
ak
vk-1
u
v2
v1
Graph Coloring
vk
ak
vl
a0
u
v
v2
v1
v
38
If P ends at vk-1, it is
with a0 since vk-1
lacked ak.
Downshifting from
vk-1, switching
colors of P, and
coloring uvk-1 with
a0 , completes the
augmentation.
ak
a0
vk-1
v2
vk
a0
a0
vl
ak
u
v1
v
ak
vk-1
v2
vk
ak
vl
ak
u
v1
v
Finally, suppose that P neither reaches vk nor vk-1, so it
ends at some vertex outside { u , vl , vk , vk-1 }
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P terminates with a0 (ak) colored edge, a case were the
far end vertex lacks ak (a0).
(The vertices along P can also touch any of u neighbors
N(u) - {vk , vk-1 }, as a0 assumed presented at those.)
We downshift from vl , give color a0 to uvl , and switch
colors of P. ■
P
a0
ak v
k-1
v2
March 2014
ak
vk
P
ak
a0
vl
ak
a0 v
k-1
u
v1
v2
v
Graph Coloring
a0
vk
ak
vl
ak
u
v1
v
40
Line Graphs
Many questions about vertices have natural analogues
involving edges.
Independent sets have no pairs of adjacent vertices;
Vertex coloring partitions the vertices
into independent sets; edges can
Definition. The line graph L(G) is
defined by V(L(G)) ≡ E(G) and ef ϵ
E(L(G)) if e = uv and f = vw , where
u,v,w ϵ V(G).
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Graph Coloring
w
G
u
f
e
v
h
g
f
L(G)
e
h
g
41
Line Graphs Characterization
Theorem. (Krausz 1943) A simple graph G is the line
graph of some simple graph iff E(G) has a partition into
cliques using each vertex of G at most twice.
Proof. Denote by H the original graph. Necessity follows
from the fact that the edges adjacent at a vertex of H
are represented in L(H) by vertices connected in a clique.
Since an edge
connects two vertices,
those vertices imply
two cliques at most.
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cliques in L(H)
N(u) u
v N(v)
vertex in L(H)
Graph Coloring
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For sufficiency, suppose E(G) has such a partition, using
cliques { S1 , … , Sk }. We shall construct H satisfying G =
L(H). Assume that G has no isolated varices.
Let v1 , … , vl be the vertices of G (if any) that appear in
a single clique of { Si }. We define a vertex in H for each
set of A = S1 , … , Sk , { v1 }, … , { vl }.
Edges of H are defined such that H vertices are adjacent
if the corresponding sets of A intersect.
Each vertex of G appears exactly in two sets of A.
Also, two vertices cannot both appear in two sets of A,
as otherwise a clique was split among the sets.
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Graph Coloring
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impossible
clique
G = L(H)
=>
H
clique
Hence there are no parallel edges in H, so it is simple,
and there is one edge in H for each vertex of G.
Adjacent vertices in G appear together in some Si and
the corresponding edges of H share the vertex
corresponding to Si. Hence G = L(H). ■
The above theorem does not directly yield an efficient
test for line graph, which the following does.
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Theorem. (Bieneke 1968) A simple graph G is the line
graph of some simple graph iff G does not contain any of
the following subgraphs as an induced subgraph.
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